Question

Differentiate the function with respect to x.

x^{sin x} + (sin x)^{cos x}

Answer 1

Let, y = x^{sin x} + (sin x)^{cos x}

Again, let y = u + v

Differentiating both sides with respect to x,

`(dy)/dx = (du)/dx + (dv)/dx`    ...(1)

Now, u = x^{sin x}

Taking logarithm of both sides,

log u = log x^{sin x}

log u = sin x log x

On differentiating both sides with respect to,

`1/u (du)/dx = sin x d/dx log x + log x d/dx sin x`

 = `sin x . 1/x + log x * cos x`

= `sin x/x + cos x log x`

`therefore (du)/dx = u (sin x/x + cos x log x)`

= `x^(sin x) (sin x/x + cos x log x)`  ....(2)

Also, v = (sin x)^{cos x}

Taking logarithm of both sides,

log v = log (sin x)^{cos x}

log v = cos x log sin x

On differentiating both sides with respect to,

`1/v (dv)/dx = cos x d/dx log sin x + log sin x d/dx cos x`

= `cos x * 1/(sin x) d/dx sin x + log sin x * (- sin x)`

= `cos x * 1/sin x * cos x - sin x log sin x`

= cos x cot x − sin x log sin x

`therefore (dv)/dx = v [cos x cot x − sin x log sin x]`

= `(sin x)^(cos x) [cos x cot x − sin x log sin x]`   ....(3)

Putting the values ​​of `(du)/dx` and `(dv)/dx` from equations (2) and (3) in equation (1), we get,

`therefore dy/dx = (du)/dx + (dv)/dx`

= `x^(sin x) (sin x/x + cos x log x) + (sin x)^(cos x) [cos x cot x − sin x log sin x]`